What does Electric Potential energy represent ,then ?
In electrostatics the electric force is conservative and so the potential energy $U$ represents the same thing that it always represents. Namely, a function of space from which the force can be derived by taking the negative of the derivative:
$$
\vec F = -\vec \nabla U\;.
$$
Or, to put it another way, in electrostatics the potential energy is something that, when combined with the kinetic energy $T$, is constant. That is, the total energy $T+U$ is a constant of the motion.
We have
$$
\vec F = -\vec \nabla U
$$
where $\vec F$ is the force due to the external electric field (the electric field sourced by all the charges that are not the test charge $q$):
$$
\vec F = q\vec E\;,
$$
where $\vec E$ is the external electric field (the electric field sourced by all the charges that are not the test charge $q$) and $q$ is the charge of the particle in the external electric field.
I.e.,
$$
q\vec E = -\vec \nabla U\;.
$$
We write:
$$
U = q\Phi
$$
such that
$$
\vec E = -\vec\nabla\Phi\;.
$$
For example, if this conservative external electric field is the only force on the particle, then it is also the net force and we can write:
$$
W_{net} = \int \vec F\cdot \vec {dx} = -\Delta U
$$
but, since in this example the electric force is the net force, then by the work-energy theorem, the above expression is also equal to the change in kinetic energy
$$
W_{net} = \Delta T
$$
and thus
$$
\Delta (T+U) = 0\;,
$$
which says that the total energy $T+U$ of a particle in a conservative force field is constant.
For example, if there is also a conservative gravitational force acting on the particle, described by a potential $U_g$ then, by a similar calculation we have:
$$
\Delta(T+U+U_g) = 0\;.
$$
So, to put it yet another way, $U$ is the usual potential energy due to the external electric field, which should be additively combined with any other potential energy functions due to any other external forces, and the kinetic energy, in order to get a constant of the motion (the total energy).
Update:
To expand this answer a little more, it might be helpful to remember where the concept of a field came from. All that experiment tells us is that if $\vec r_0$ and $\vec r_1$ are two fixed points in space where two charges, $q_0$ and $q_1$, respectively, are located then the force that $q_0$ feels due to $q_1$ is:
$$
\vec F_0 = \frac{q_0q_1(\vec r_0 - \vec r_1)}{|\vec r_0 - \vec r_1|^3}\;.
$$
Similarly, if there are charges $q_0$, $q_1$, $q_2$, and $q_3$, the force felt by $q_0$ is:
$$
\vec F_0 = \frac{q_0q_1(\vec r_0 - \vec r_1)}{|\vec r_0 - \vec r_1|^3}
+\frac{q_0q_1(\vec r_0 - \vec r_1)}{|\vec r_0 - \vec r_1|^3}
+\frac{q_0q_1(\vec r_0 - \vec r_1)}{|\vec r_0 - \vec r_1|^3}
+\frac{q_0q_1(\vec r_0 - \vec r_1)}{|\vec r_0 - \vec r_1|^3}
$$
$$
=q_0\sum_{j=1}^3\frac{q_j(\vec r_0 - \vec r_j)}{|\vec r_0 - \vec r_j|^3}\;,
$$
where, we are considering the force on $q_0$, so here I will call $q_0$ the "test charge."
Similarly, if there are $N+1$ charges $q_0$, $q_1$, $q_2$, ..., $q_N$, the force felt by $q_0$ is:
$$
\vec F_0=q_0\sum_{j=1}^N\frac{q_j(\vec r_0 - \vec r_j)}{|\vec r_0 - \vec r_j|^3}\;,
$$
Clearly, the force on the test charge can be written as a vector field evaluated at $\vec r_0$:
$$
\vec F_0 = \vec F(\vec x = \vec r_0)\;,
$$
where
$$
\vec F(\vec x) = q_0\sum_{j=1}^N\frac{q_j(\vec x - \vec r_j)}{|\vec x - \vec r_j|^3}\;.\tag{A}
$$
This seems like a minor change, but the introduction of the field (which is what physicists call a function of space) is important conceptually.
But now our Eq. (A) looks a little funny... it has lost its dependence on $\vec r_0$, but it still depends on $q_0$. This might motivate us to consider a different field, the field in Eq. (A) divided by the test charge. We call this field the "electric field."
$$
\vec E(\vec x) = \sum_{j=1}^N\frac{q_j(\vec x - \vec r_j)}{|\vec x - \vec r_j|^3}\;,
$$
which doesn't depend on the test charge ($q_0$) or position ($\vec r_0$) explicitly, but does depend on the distribution of all the other "external" charge in the world ($q_1$, $q_2$, ..., $q_N$).
In terms of potentials, we can write the force in Eq. (A) as:
$$
F(\vec x) = -\vec \nabla U\;,
$$
where
$$
U(\vec x) = q_0\sum_{j=1}^N\frac{q_j}{|\vec x - \vec r_j|}+C\;,
$$
where $C$ is an arbitrary constant, which we here choose to be $C=0$.
This means that the Work done by the source charges on the test charge $q_0$ as it moves from infinity (infinitely far away from the source charges) to $\vec r_0$ is:
$$
W = \int \vec F\cdot\vec{dx} = -\Delta U = -q_0\sum_{j=1}^N\frac{q_j}{|\vec r_0 - \vec r_j|} = -U(\vec r_0)\;.
$$
We see that, as usual, the work done by the external force due to the source charges is the negative of the change in potential energy $U$ due to the source charges. Here the test charge $q_0$ again only comes in as a single overall multiple, which perhaps motivates the definition of the electric potential $\Phi$ as:
$$
\Phi(\vec x) = \frac{U}{q_0} = \sum_{j=1}^N\frac{q_j}{|\vec x - \vec r_j|}\;.
$$
Or, again in terms of the field:
$$
\vec F_0 = q_0\vec E_0\;,
$$
where
$$
E_0 = E(\vec x=\vec r_0) = \sum_{j=1}^N\frac{q_j(\vec r_0 - \vec r_j)}{|\vec x - \vec r_j|^3}\;.
$$
Here, we see that the force acting on the test charge $q_0$ is equal to the electric field of all the other charges in the world (not including $q_0$ itself) multiplied by the test charge value $q_0$.
[Question in comments:] "the gradient of electric potential energy... tells us how much force a charge Q, a non unit charge , would experience... what happens to electric field of charge Q itself ,is it considered ?"
The charge $Q$ itself is not considered as one of the sources that created the electric field $\vec E$ acting on $Q$. It is only considered in the sense that you still have to multiply the electric field by a single overall factor of $Q$ to get the force on $Q$.
